(1 point) A tank contains 1080 L of pure water. Solution that contains 0.05 kg o
ID: 2856395 • Letter: #
Question
(1 point) A tank contains 1080 L of pure water. Solution that contains 0.05 kg of sugar per liter enters the tank at the rate 7 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate.
(a) How much sugar is in the tank at the begining?
y(0)= (kg)
(b) Find the amount of sugar after t minutes.
y(t)= (kg)
(c) As t becomes large, what value is y(t) approaching ? In other words, calculate the following limit. limty(t)= (kg)
Note : Please show clearly the final answer
Explanation / Answer
(a). 0 kg
(b).
Let Q is mass of sugar after t minutes.
dQ/dt = (0.05)(7) - (7/1080)Q
rearrange diff. eqs.
dQ/dt + (7/1080)Q = 0.35
integrating factor,
f = e^( (7/1080) dt) = e^((7/1080)t)
Q f = 0.35 f dt
Q e^((7/1080)t) = 0.35 e^((7/1080)t) dt
Q e^((7/1080)t) = 0.35 * (1080/7) e^((7/1080)t) + C
initial condition
Q(0) = 0
0 = 0.35 * (1080/7) e^((7/1080) * 0) + C
C = 54
general solution is
Q e^((7/1080)t) = 54 e^((7/1080)t) + 54
Q(t) = 54(1 - e^(-(7/1080)t))
(c).
for t infinity
Q() = 54 (1 - e^(-(6/2900)* ))
Q() = 54 kg
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